- Thread starter Elix
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\(\displaystyle A=2\sqrt{2}\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)(\sin(\theta)+1)^{\frac{3}{2}}\,d\theta\)

Now it appears the integration should be straightforward.

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\(\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta\)

And our definite integral becomes:

\(\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}\)

Thanks a bunch! I struggled to connect how you got to that simplification at first but figured it out. I really appreciate your help with this and my previous question!

\(\displaystyle u=\sin(\theta)+1\implies du=\cos(\theta)\,d\theta\)

And our definite integral becomes:

\(\displaystyle A=2\sqrt{2}\pi\int_0^2 u^{\frac{3}{2}}\,du=2\sqrt{2}\pi\left(\frac{2}{5}2^{\frac{5}{2}}\right)=\frac{32\pi}{5}\)